Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

The bonds present in N_(2)O_(5) are

Answer»

IONIC
COVALENT
Ionic and covalent
Covalent and DATIVE

ANSWER :D
Discussion
2.

The bond present in NaNC are

Answer»

IONIC BOND
COVALENT bond
Co-ordinate covalent bond
All

Answer :D
Discussion
3.

The bond order of NO_(3)^(-)ion is ………… and its shape is ……….

Answer»

SOLUTION :1.33 PLANER TRIANGULAR
Discussion
4.

The bond order of O_(2)^(-) is

Answer»

2
`1.5`
1
`2.5`

SOLUTION :(B) `BO=(1)/(2)(6-3)=1.5`
Discussion
5.

The bond order ofN_(2), N_(2)^(+) , N_(2) ^(-) andN_(2)^(2-) respectively are ……….., …………..,…………..and…………….

Answer»

SOLUTION :3, 2.5 , 2.5 , 2
Discussion
6.

The bond order of He^(2+) ion is-

Answer»

0
0.5
1
1.5

Solution :There is no electron present in `He^(2+)` ION. THUS BOND order=0.
Discussion
7.

The bond order of N-O bonds in NO_(3)^(-)ion is

Answer»

`0.33`
`1.00`
`1.33`
`1.50`

SOLUTION :
`(2 + 1 + 1 )/(3) = 1.33` .
Discussion
8.

The bond order of individual carbon bonds in benzene is

Answer»

ONE
TWO
Between one and two
NONE of these

ANSWER :C
Discussion
9.

The bond order of CO molecule on the basis of MO theory is

Answer»

Zero
2
1
3

Answer :D
Discussion
10.

The bond order of a molecule is given by

Answer»

the DIFFERENCE between the NUMBER of ELECTRONS in BONDING and antibonding orbitals.
total number of electrons in bonding and antibonding orbitals.
Twice the difference between the number of electrons in bonding and antibonding orbitals.
Half the difference between umbe of electrons in bonding and antibonding orbitals.

Answer :D
Discussion
11.

The bond order in O_(2)^(2-) species is

Answer»

1
2
3
4

Answer :A
Discussion
12.

The bond order in O_(2)^(-) species is

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1
2
3
4

Answer :A
Discussion
13.

The bond order

Answer»

Can have NEGATIVE VALUE
Is any NUMBER other than ZERO
Is any integer
Can have any value INCLUDING zero

Answer :D
Discussion
14.

The bond length of two N-O bond in nitromethane is…..

Answer»

It is a between of SINGLE bond N-O and double bond N= O
It is a single bond `N-O` as WELL as double bond N= O
It is a half of the summation of single bond `N- O` and double bond N= O
Both (A) and (C )

ANSWER :D
Discussion
15.

The bond length of H_(2) , F_(2)and HF is 74 pm, 144 pm and 92 pm respectively. Which one is most stable ?

Answer»

`F_(2)`
`H_(2)`
HF
All are EQUAL

SOLUTION :`H_(2)`
Discussion
16.

The bond length is affected by

Answer»

Hybridisation
Delocalisation
Electronegativity
All of the above

Answer :D
Discussion
17.

The bond length between central carbon atom and other carbon atom is minimun in

Answer»

Propene
Propyne
Propane
Pentane

Answer :B
Discussion
18.

The bond length between C2 and C3 in acrutl aldehyde is not equal to the bond length between carbons of ethene because in acryl aldehydedouble is in conjugation , so it it shows resonance which results in an increase in the bond length betweenC2 and C3 in acryl aldehyde . Find out the correct statement (s) about the given compound ?

Answer»

Bond length between C2 and C3 = Bond length between C5 and C6 .
Bond length between C1 and C2 = Bond length between C6 and C7.
Bond length between C6 and C7 `lt` Bond length between C3 and C4.
Bond length between C2 and C3 `lt` Bond length between C3 and C4

ANSWER :C
Discussion
19.

The bond length between C2 and C3 in acrutl aldehyde is not equal to the bond length between carbons of ethene because in acryl aldehydedouble is in conjugation , so it it shows resonance which results in an increase in the bond length betweenC2 and C3 in acryl aldehyde . Which compound does not have the conjugative system to show the resonance ?

Answer»

`CH_(2)=CH-CH=CH-CH_(2)-CH_(3)`


ANSWER :D
Discussion
20.

The bond length between C2 and C3 in acrutl aldehyde is not equal to the bond length between carbons of ethene because in acryl aldehydedouble is in conjugation , so it it shows resonance which results in an increase in the bond length betweenC2 and C3 in acryl aldehyde . Which compound will not show the resonance ?

Answer»




ANSWER :B
Discussion
21.

The bond length between all the four carbon atoms is same in 1, 3 - butadine. Explain with reason.

Answer»

SOLUTION :`CH_(2)=CH-CH=CH_(2) harr overset(+)CH_(2)-CH=CH-overset(bar..)CH_(2)`
`CH_(2)bar(...)CH bar(...)CHbar(...)CH_(2)-=overset(bar(..))CH_(2)-CH=CH-overset(+)CH_2`
(Partial double bond character)
`1,3 ` ,3-butadiene is a conjugated MOLECULE with four overlapping p-orbital on adjacent atoms. And A-electrons are delocalised over four atoms. This SHORTENS the bond length of central C bond Thus, the bond length between all the four C-atoms are same in 1,3-butadiene.
Discussion
22.

Given that bond energies of N=N,H-H and N-H bonds as 945, 436 and 391 kJ/ mol respectively ,the enthalpy of the reaction N_(2)(g)+3H_(2)(g)+to2NH_(3)(g),is

Answer»


ANSWER :`-93KJ`
Discussion
23.

The bond enthalpies of D-D and O-O and D-O are respectively +440 ,+498 and +491.5KJ "mol"^(-1) calculation Delta H for the reaction D_(2)(g) +1/2O_(2)(g)to D_(2) O(l)

Answer»

SOLUTION :Heat of REACTION, =(Enthalpy of bonds DISSOCIATED )-(Enthalpy of bonds formed)
`DELTA H=(440+498/2)-(2xx491.5)=689-983=-294KJ`
Enthalpy of the given reaction is -294KJ.
Discussion
24.

The bond enthalpies of D-Dand O-O and D-O are respectively, + 440, +498 and + 491.5 kJ mol^(-1). Calculate DeltaH for the reaction D_(2(g)) + 1/2 O_(2(g)) to D_2O_((l))

Answer»

SOLUTION :Heat of reaction, `DeltaH `= (ENTHALPY of bonds dissoicated) - (Enthalpy of bonds FORMED)
`DeltaH = (440 + 498/2) - (2 xx 491.5)`
`= 689 - 983 = -294 kJ`
Enthalpy of the GIVEN reaction is -294 kJ.
Discussion
25.

The bond enthalpies of D-D and 0-0 and D - ( are respectively, + 440, + 498 and + 491.5 kJ mol^(-1). Calculate DeltaH for the reaction D _(2 (g)) + (1)/(2) O _(2(g)) to D _(2) O _((l))

Answer»

SOLUTION :Heat of reaction, `DeltaH = `(Enthalpy of bonds dissociated) - (Enthalpy of bonds formed)
`Delta H = (440 + (498)/(2)) - (2 xx 491 .5)`
`=689-983 =-294 KJ`
Enthalpy of the GIVEN reaction is `- 294 kJ .`
Discussion
26.

The bond energy of H_(2) is foundto be 435 kJ mol^(-1). Therefore, the enthalpy of formation of hydrogen atom will be "…................"

Answer»

SOLUTION : `435 //2 = 217.5 KJ MOL^(-1)`
Discussion
27.

The bond energy of H_(2(g)) is 436kJ/mole. The bond enthalpy of N_(2(g)) is 941.3 kJ/mole. What is the bond enthalpy of N - H if NH_3 formation energy is -46.0 kJ/mole ?

Answer»

SOLUTION :390 kJ/mole
Discussion
28.

The bond energy of an O-H bondis 109 kcal // mol. When a mole of water is formed,then

Answer»

109 kcal is released
218 kcal is released
109 kcal is absorbed
218 kcal is absorbed

Solution :Bond ENERGY of O-H bond `=109 kcal // `mol means that whn 1 mole of O-H bondsare formed,109 kcal energy is released ( because dissociationof bond absorbs energy ). Formation of one mole of WATER is equivalent to formation of2moles of O-H bond ( H-O-H) . Hence,energy released `=2 xx 109=218 kcal`. ( Strictly SPEAKING, we CALCULATE`Delta_(f)H^(@) `for the reaction.
`H_(2)(g)+(1)/(2) O_(2)(g) rarr H_(2)O(L) .i.e., ( H-O-H)`
`Delta_(f)H^(@) =BE(H-H) + (1)/(2) BE(O=O)-2BE(O-H)`
Discussion
29.

The bond energy depends upon _____

Answer»

1) SIZE of the atom
2) ELECTRONEGATIVITY
3) BOND length
4) all of the above

Answer :D
Discussion
30.

The bond energies of H - H, X - X and H - X are 104 K.Cal, 38K.Cal and 138 K.Cal respectively the electronegativity of 'X' is [sqrt(67) = 8.18]

Answer»

3
3.5
3.8
1.7

Answer :C
Discussion
31.

The bond dissoication energies of X_(2), Y_92) and XY are in the ratio of 1: 0.5 : 1 . DeltaH for the formation of XY is -200 kJ mol^(-1). The bond dissociation energy ofX_(2) will be

Answer»

`200 kJ mol^(-1)`
`100 kJ mol^(-1)`
` 800 kJ mol^(-1)`
` 400 kJ mol^(-1)`

Solution :Suppose the BOND dissociation energy of`X_(2) = 'a' kJ mol^(-1), i.e., BE(X_(2))= akJmol^(-1)`then `BE(Y_(2)) =0.5 a` andBE of XY `=akJmol^(-1)`
Given `=(1)/(2) X_(2)+(1)/(Y_(2)) rarr XY ,DELTAH = - 200 kJ mol^(-1)`
`Delta_(r) H=` BE (Reactants) - BE( Products)
`= [ (1)/(2) BE(X_(2))+(1)/(2) BE(Y_(2))]-BE(XY)`
`:. - 200 = (a)/(2)+ ( 0.5)/( 2) -a =0.5a+ 0.25a-a =-0.25 a`
`:. a=( 200)/( 0.25) = 80 kJ mol^(-1)`
Discussion
32.

The bond dissociation of the molecules A_2 , B_2,C_2 are 498, 158, 945 kJ/ mole respectively. If so, the correct decreasing order of their bond orders is

Answer»

`A_2 ,B_2 ,C_2`
`C_2 ,B_2 ,A_2`
`C_2 ,A_2 ,B_2`
`B_2,C_2 ,A_2`

ANSWER :C
Discussion
33.

The bond dissociation energy of methane and ethane are 360 kJ "mol"^(-1) and 620 kJ "mol"^(-1) respectively. Then, the bond dissociation energy of C-C bond is _____

Answer»

`170 KJ "mol"^(-1)`
`50 KJ "mol"^(-1)`
`80 KJ "mol"^(-1)`
`220 KJ "mol"^(-1)`

Solution :4(C-H)is equal to 360KJ/mol
(C-H) is equal to 90 KJ/mol
Discussion
34.

The bond dissociation energy of B-F in BF_(3) is 646 kJ "mol"^(-1) whereas that of C-F in CF_(4)is 515 kJ "mol"^(-1).The correct reasonfor higherB-Fbond dissociationenergy as compared to thatof C-F is

Answer»

stronger `sigma`-bond between B and F in `BF_(3)` as compared to that between `C` and `F` in `CF_(4)`
significant `ppi-ppi`interactionbetween B and F in `BF_(3)` whereas there is no POSSIBILITY of suchinteraction between C and F in `CF_(4)`
LOWER degree of `ppi-ppi`interactionbetween B and F in `BF_(3)` than that between C and F in `CF_(4)`.
smaller size of B-atom as compared to that of C-atom

Solution :Due to identical size of 2p-orbitalson B and F,back donation of a LONE PAIR of electrons from the filled 2o-orbitals of F to the empty 2p-orbital of B occurs to a considerable extent. As a result , `B-F`bond has some double character and hencebond dissociationenergy of `B-F`bond in `BF_(3)`is much higherthan that of C-F bondin `CF_(4)`.the reasonbeingthat C does not have an empty 2p-orbitals and hence back donation does notoccur.
Discussion
35.

The bond dissociation energy needed to form the benzyl radical from toluene is................than the energy needed for the formation of methyl radical from methane.

Answer»


ANSWER :LOWER
Discussion
36.

The bond dissociation energy of B - F in BF_(3) is 646 KJ mol^(-1) whereas that of C-F in CF_(4) is 515 kJ mol^(-1) . The correct reason for higher B - F bond dissociation energy as comparad to that of C - F is

Answer»

smallarsize of B-atom as compared to that of
C-atom
STRONGER `sigma`-bond between B and F in `BF_(3)`as
compared to that between C and F in `CF_(4)`
significant `ppi-ppi` interaction between B and
F in`BF_(3)` whereas there is no possibility of such
interaction between C and F in `CF_(4)`
lower degree of `ppi-ppi` interaction between B
and F in `BF_(3)` than between C and F in
`CF_(4) `

Solution :The empty `2p_(z)` orbitals of B which is not involved
in the hybridisation can accept an electron pair
from a FULL `2p_(z)` orbitals of any F-atom , formaing
` pi`-bond . Thus , B-F bondlength becomes shorter.
HENCE, bond is stronger and dissociation energy
is higher .
Discussion
37.

The bond dissociation energy depends upon the nature of the bond and nature of the molecule. If any molecule more than 1 bonds of similar nature are present then the bond energy reported is the average bond energy. If enthalpy of hydrogenation of C_(6)H_(6(l)) "into" C_(6)H_(12(l)) " is " -205 kJ and resonance energy of C_(6)H_(6(l)) is -152kJ//mol then enthalpy of hydrogenation of is ? Answer Delta H_("vap") " of " C_(6)H_(6(l)), C_(6)H_(10(l)), C_(6)H_(12(l)) all are equal:

Answer»

`-535.5` kJ/mol
`-238` kJ/mol
`-357` kJ/mol
`-119` kJ/mol

Solution :Heat of HYDROGENATION of cyclo HEXANE
`=("Heat of Hydrogenation of Benzene + Resonance ENERGY")/(3)`
`= (-205 + (-152))/(3) = -199kJ`
Discussion
38.

The bond dissociation energy depends upon the nature of the bond and nature of the molecule. If any molecule more than 1 bonds of similar nature are present then the bond energy reported is the average bond energy. Determine C-C and C-H bond enthalpy (in kJ/mol). Given: Delta_(f)H^(0) (C_(2)H_(6),g)= -85kJ//mol, Delta_(f) H^(0) (C_(3)H_(8), g)= -104kJ//mole, Delta_("sub")H^(0) (C,s)= 718kJ//mol, B.E. (H-H)= 436 kJ/mol,

Answer»

414345
345414
287405.5
none of these

Solution :The given reactions are `2C_(("graphite")) + 3H_(2(g)) rarr C_(2)H_(6(g)), Delta H = -85 kJ`/mole …(1)
`3C_(("graphite")) + 4H_(2(g)) rarr C_(3)H_(8(g)), Delta H = - 104` kJ/mole …(2)
`C_(("graphite")) rarr C_((g)), Delta H = 718` kJ/mole …(3)
`H_(2) rarr 2H, Delta H = 436` kJ/mole ...(4)
To get expression for heat of reaction in terms of Bond ENERGY we have to use the expression with all the components are in gaseous state. So, from Eq. (1) and (3)
`2C_((g)) + 3H_(2(g)) rarr C_(2)H_(6(g)), Delta H = - 1521 kJ` ...(5)
From Eq.(2) and (3)
`3C_((g)) + 4H_(2(g)) rarr C_(3)H_(8(g)) , Delta H = - 2258kJ` ...(6)
We know that heat of reaction `Sigma` B.E of reactants `= Sigma` B.E of products. So,
`-1521= 3 (H- H) - [6 C-H + C-C]` ...(7)
`-2258 = 4(H-H) - [8C- H + 2C -C]` ..(8)
On solving the above two equations. C- H bond energy = 414
C- C bond energy = 345
Discussion
39.

The bond dissociation energies of X_(2), Y_(2) and XY are in the ratio of1 : 0.5 : 1. DeltaH for the formation of XY is -200 kJ mol^(-1).The bond dissociation energy of X_(2) will be

Answer»

`200 kJ"MOL"^(-1)`
`100 kJ"mol"^(-1)`
`800 kJ"mol"^(-1)`
`400 kJ"mol"^(-1)`

Solution :Let bond DISSOCIATION energy of `X_(2)` be x kJ `mol^(-1)`
Then
B.E. Of `Y_(2) = 0.5 x kJ mol^(-1)`
B.E of XY = x kJ `mol^(-1)`
`1/2 X_(2) + 1/2Y_(2) to XY DeltaH = -200 kJ mol^(-1)`
`DeltaH = Sigma`B.E (reactants) - `Sigma`B.E(PRODUCTS)
`DeltaH = 1/2 B.E.(X_(2)) + 1/2 B.E. (Y_(2)) - B.E.(XY)`
` -200 = 1/2(x) + 1/2 (0.5 x) - 1(x)`
`-200 = -0.25x`
`therefore x = (200)/(0.25) = 800 kJ"mol"^(-1)`.
Discussion
40.

The bond dissociation energies of X_2, Y_2 and XY are in the ratio of 1: 0.5 : 1. Delta Hfor the formation of XY is -200 kJ "mol"^(-1) . The bond dissociation energy of X_2will be

Answer»

`400 kJ "mol"^(-1)`
`200 kJ "mol"^(-1)`
`800 kJ "mol"^(-1)`
`100 kJ "mol"^(-1)`

Solution :`800 kJ "mol"^(-1)`
Discussion
41.

The bond dissociation energies of gaseous H_(2), Cl_(2) and HCl are 104, 58 and 103 kCal/mole respectively. If magnitude of heat of formation of HCl = 11x, x=?

Answer»


Solution :Reaction for formation of HCl is `(1)/(2) H_(2(g)) + (1)/(2) Cl_(2(g)) rarr HCl_((g))`
As all are in gaseous state
`DELTA H = Sigma` B.E of Reactions `- Sigma` B.E. of PRODUCTS
`= (1)/(2) (104) + (1)/(2) (58) - 103 = -22`
Discussion
42.

The bond dissociation energies forCl_(2),I_(2) and ICIare 242.3,151.0 and211.3 kJ //mole respectively . The enthalpyof sublimation of iodine is62.8 kJ // mole.What is the standard enthalpy of formation ofICI(g) nearly equal to ?

Answer»

`- 211.3 kJ //`MOLE
`- 14.6kJ //`mole
16.8 kJ`//` mole
`33.5 kJ//` mole

Solution :We aim `:(1)/(2) I_(2)(s) + (1)/(2) Cl_(2) (g) RARR ICI (g) `
We are given`:`
(i) `Cl_(2)(g)rarr 2Cl(g), DeltaH=242.3 kJ`
(II) `I_(2)(g) rarr 2I(g),DeltaH = 151 kJ`
(iii) `ICI (g) rarr (i) + Cl(g), DeltaH =211.3 kJ`
(iv) `I_(2)(s) rarrI_(2)(g), DeltaH =62.8 kJ`
`(1)/(2) (ii) (1)/(2) (i)- (iii) +(1)/(2) ` gives the required EQUATION with
`DeltaH =(1)/(2) (151) +(1)/(2) (292.3) -211.3 + (1)/(2) (62.8) = 16.75kJ`
Discussion
43.

The bond dissociatio of the molecules A_(2),B_(2),C_(2) are 498, 158, 945 kJ/mole respectively. If So, the correct decreasing order of their bond orders is

Answer»

`A_(2),B_(2),C_(2)`
`C_(2),B_(2),A_(2)`
`C_(2),A_(2),B_(2)`
`B_(2),C_(2),A_(2)`

ANSWER :C
Discussion
44.

The bond between two identical non-metal atoms has a pair of electrons

Answer»

unequally SHARED between the TWO
TRANSFERRED fully from one ATOM to another
with identical spin
equally shared between them

Answer :D
Discussion
45.

The bond between chlorine and bromine in BrCl is

Answer»

ionic
non-POLAR
polar with NEGATIVE END on Br
polar with negative end on CL

Answer :D
Discussion
46.

The bond between carbon atom (1) and carbon atom (2) in the compound N -= overset(1)(C )-overset(2)(C )H= overset(3)(C )H_(2) invoves the hybrids as…..

Answer»

`SP and sp^(2)`
`sp^(2) and sp^(2)`
`sp and sp`
`sp^(3) and sp`

ANSWER :A
Discussion
47.

The bond between carbon atom (1) and carbon atom (2) in compoundN-=overset(1)C-overset(2)CH=CH_2involves the hybridization 1) spa and sp?

Answer»

`SP^(2)" and "sp^(2)`
`sp^(3)" and "sp`
`sp" and "sp^(2)`
`sp" and "sp`

ANSWER :C
Discussion
48.

The bond between B and C will be

Answer»

Ionic
Colvalent
Hydrogen
Coordinate

Solution :The bonds formed between B and C i.e. in `PCl_(5)` will be COVALENT in nature.
Discussion
49.

The bond between B and C will be ...

Answer»

ionic
COVALENT
hydrogen
coordinate

SOLUTION :covalent
The bond between B and C will be covalent. Both Band Care non-metal atoms. B represents PHOSPHORUS and C represent chlorine.
Discussion
50.

The bond angle in H_2 O is 105^@but in H_2 S is 92^@ . Explain the difference.

Answer»

Solution :The larger size of the S atom compared with O atom, minimises the elcectron REPULSIONS and allows the BONDS in `H_(2)S` to be more purely p-type.
Moreover O in `H_(2)O` is `sp^(3)` hybridised but pure p-orbitals are USED by S of `H_(2)S`
Discussion